# Implement a Non-Guessing Sudoku Solver

Implement the shortest Sudoku solver.

Sudoku Puzzle:

 | 1 2 3 | 4 5 6 | 7 8 9
-+-----------------------
A|   3   |     1 |
B|     6 |       |   5
C| 5     |       | 9 8 3
-+-----------------------
D|   8   |     6 | 3   2
E|       |   5   |
F| 9   3 | 8     |   6
-+-----------------------
G| 7 1 4 |       |     9
H|   2   |       | 8
I|       | 4     |   3


 | 1 2 3 | 4 5 6 | 7 8 9
-+-----------------------
A| 8 3 2 | 5 9 1 | 6 7 4
B| 4 9 6 | 3 8 7 | 2 5 1
C| 5 7 1 | 2 6 4 | 9 8 3
-+-----------------------
D| 1 8 5 | 7 4 6 | 3 9 2
E| 2 6 7 | 9 5 3 | 4 1 8
F| 9 4 3 | 8 1 2 | 7 6 5
-+-----------------------
G| 7 1 4 | 6 3 8 | 5 2 9
H| 3 2 9 | 1 7 5 | 8 4 6
I| 6 5 8 | 4 2 9 | 1 3 7


Rules:

1. Assume all sudokus are solvable by logic only.
2. All input will be 81 characters long. Missing characters will be 0.
3. Output the solution as a single string.
4. The "grid" may be stored internally however you wish.
5. The solution must use a non-guessing solution. (see Sudoku Solver)

Example I/O:

>sudoku.py "030001000006000050500000983080006302000050000903800060714000009020000800000400030"
832591674496387251571264983185746392267953418943812765714638529329175846658429137

• You should really add a time limit. Commented Feb 2, 2011 at 0:30
• @JPvdMerwe: Good point, but a time limit would be hard to standardize. Commented Feb 2, 2011 at 1:19
• @gnibbler: It might have been done before (but not on codegolf.se). I think it will still be fun to solve as well as add some value to the community, especially if one goes about it honestly. Commented Feb 2, 2011 at 1:47
• I like this one. I've been hesitant to try an actual golf solution, and I've been thinking about writing a Sudoku solver (it seems like a fun exercise). I think it's something people like me, who've never golfed before, could use as a jumping-off point. And once I come up with one, I might then golf it.
– Andy
Commented Feb 2, 2011 at 4:47
• Problems "solvable by logic only" is very vague. Do you mean, perhaps, using only the basic steps of a) Writing a value in a cell for which it's value not in its row, column, and block b) Identifying a number that can only go in one place in its row, column, or block, and writing it there?
– xnor
Commented May 17, 2014 at 2:41

RUBY (449 436 chars)

I=*(0..8)
b=$*[0].split('').map{|v|v<'1'?I.map{|d|d+1}:[v.to_i]};f=b.map{|c|!c[1]} [[z=I.map{|v|v%3+v/3*9},z.map{|v|v*3}],[x=I.map{|v|v*9},I],[I,x] ].map{|s,t|t.map{|i|d=[a=0]*10;s.map{|j|c=b[i+j];c.map{|v|d[v]+=1if !f[i+j]} v,r=*c;s.map{|k|b[i+k].delete(v)if j!=k}if !r s[(a+=1)..8].map{|k|s.map{|l|b[i+l]-=c if l!=k&&l!=j}if c.size==2&&c==b[i+k]}} v=d.index 1;f[i+k=s.find{|j|b[i+j].index v}]=b[i+k]=[v]if v}}while f.index(!1) p b*''  Example: C:\golf>soduku2.rb 030001000006000050500000983080006302000050000903800060714000009020000800000400030 "832591674496387251571264983185746392267953418943812765714638529329175846658429137"  quick explanation: Board b is an array of 81 arrays holding all the possible values for each cell. The array on line three holds [offset,start_index] for each group (boxes,rows,columns). Three tasks are performed while iterating through the groups. 1. The value of any cell of size 1 is removed from the rest of the group. 2. If any pair of cells contain the same 2 values, these values are removed from the rest of the group. 3. The count of each value is stored in d - if there is only 1 instance of a value, we set the containing cell to that value, and mark the cell fixed in f Repeat until all cells are fixed. • You can omit the brackets in I=*(0..8), will save 2 chars. Commented Feb 8, 2011 at 17:02 • I get sudokusolver.rb:8: unterminated string meets end of file if I start it with ruby1.8 sudokusolver.rb 030.... What am I doing wrong? Commented May 4, 2011 at 23:53 • Looks like there is an extra ' on the last line. Not sure how that got there... Commented May 5, 2011 at 2:27 # Prolog - 493 Characters :-use_module(library(clpfd)). a(X):-all_distinct(X). b([],[],[]). b([A,B,C|X],[D,E,F|Y],[G,H,I|Z]):-a([A,B,C,D,E,F,G,H,I]),b(X,Y,Z). c([A,B,C,D,E,F,G,H,I|X])-->[[A,B,C,D,E,F,G,H,I]],c(X). c([])-->[]. l(X,Y):-length(X,Y). m(X,Y):-maplist(X,Y). n(L,M):-l(M,L). o(48,_). o(I,O):-O is I-48. :-l(L,81),see(user),m(get,L),seen,maplist(o,L,M),phrase(c(M),R),l(R,9),m(n(9),R),append(R,V),V ins 1..9,m(a,R),transpose(R,X),m(a,X),R=[A,B,C,D,E,F,G,H,I],b(A,B,C),b(D,E,F),b(G,H,I),flatten(R,O),m(write,O).  Output: Inputting: 000000000000003085001020000000507000004000100090000000500000073002010000000040009 Outputs: 987654321246173985351928746128537694634892157795461832519286473472319568863745219 Inputting: 030001000006000050500000983080006302000050000903800060714000009020000800000400030 Outputs: 832591674496387251571264983185746392267953418943812765714638529329175846658429137 # Perl 5, 295 bytes sub P{$r=$p/9;$c=$p%9;$q=3*~~($r/3)+$c/3}sub N{($i,$u)=@_;P;$G[$p]=$u*$i;$R[$r]{$i}=$C[$c]{$i}=$Q[$q]{$i}=$u}sub S{if($p>80){print@G;exit}if($G[$p]){S(++$p);$p--}else{N($_,1),S(++$p),$p--,N($_)for grep{P;$R[$r]{$_}+$C[$c]{$_}+$Q[$q]{$_}<1}1..9}}@G=split'',$ARGV[0];N($G[$p=$_],1)for 0..80;$p=0;S  Try it online! Shortened from 339 to 317 to 315 to 314 to 310 to 309 to 298 to 295 • nice. there is the test case in the Prolog answer 000000000000003085001020000000507000004000100090000000500000073002010000000040009. Would you like to run your solution with this test case? Commented Jul 15, 2020 at 10:47 • Nice example :-) . It is kind of worst case, with a solution having 987654321 in the first line. It runs into the time limit on TIO, but run locally on my PC it still outputs the correct solution -- after around 10 minutes. There is, of course, room for performance optimization, but it would make the code longer. Commented Jul 15, 2020 at 14:05 • Thanks. This is CodeGolf, so a performance does not matter Commented Jul 15, 2020 at 15:35 • I'm afraid the grep uses guesswork and your solution uses a brute force. I'm guessing :) your answer would be better suited for Implement a Brute Force Sudoku Solver. See rule 5 in this question. Commented Jul 15, 2020 at 16:22 • If backtracking were not allowed, a solution in Prolog should not be valid either because Prolog does backtracking inherently. It is just hidden under the surface. Commented Jul 15, 2020 at 16:41 # PowerShell, 491490488465 463 bytes $w=,0*81;$c=,(1..9)*81$a=(24,24,6)*3|%{,(0..8|%{($r++)});,(0..8|%{$v%81;$v+=9});$v++;,((1,1,7)*3|%{+$q;$q+=$_});$q-=$_} filter g($v,$i){if($v){$a|?{$i-in$_}|%{$_|%{$c[$_]=$c[$_]-ne$v}};$w[$i]=$v;$c[$i]=@()}}
$args|% t*y|%{g($_-48)($i++)} for(;$a|%{($b=$_)|?{($v=$c[$_])}|?{(($c[$b]|%{"$_"})-eq$v).Count-eq$v.Count
$c[$b]|%{$_}|group|? C* -eq 1|? N* -in$v|%{$v=+$_.Name;1}}|%{$b|?{"$v"-ne$c[$_]}|%{$c[$_]=@($c[$_]|?{$_-notin$v})}
g($v[0]*!$v[1])$_ 1}}){} -join$w


Try it online!

Implemented Solving methods: Single, Naked Pair, Naked Trio, Naked Quad and Hidden Single.